Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $t \neq 0$. $r = \dfrac{-5t^2 - 35t}{-7t + 63} \times \dfrac{t - 4}{t^2 + 3t - 28} $
Answer: First factor the quadratic. $r = \dfrac{-5t^2 - 35t}{-7t + 63} \times \dfrac{t - 4}{(t + 7)(t - 4)} $ Then factor out any other terms. $r = \dfrac{-5t(t + 7)}{-7(t - 9)} \times \dfrac{t - 4}{(t + 7)(t - 4)} $ Then multiply the two numerators and multiply the two denominators. $r = \dfrac{ -5t(t + 7) \times (t - 4) } { -7(t - 9) \times (t + 7)(t - 4) } $ $r = \dfrac{ -5t(t + 7)(t - 4)}{ -7(t - 9)(t + 7)(t - 4)} $ Notice that $(t - 4)$ and $(t + 7)$ appear in both the numerator and denominator so we can cancel them. $r = \dfrac{ -5t\cancel{(t + 7)}(t - 4)}{ -7(t - 9)\cancel{(t + 7)}(t - 4)} $ We are dividing by $t + 7$ , so $t + 7 \neq 0$ Therefore, $t \neq -7$ $r = \dfrac{ -5t\cancel{(t + 7)}\cancel{(t - 4)}}{ -7(t - 9)\cancel{(t + 7)}\cancel{(t - 4)}} $ We are dividing by $t - 4$ , so $t - 4 \neq 0$ Therefore, $t \neq 4$ $r = \dfrac{-5t}{-7(t - 9)} $ $r = \dfrac{5t}{7(t - 9)} ; \space t \neq -7 ; \space t \neq 4 $